例:
// result == true; Oct 23,2010 is a Saturdayvar result = InBetweenDaysInclusive(new DateTime(2010,10,23),DayOfWeek.FrIDay,DayOfWeek.Sunday);// result == true; Oct 22,2010 is a FrIDayresult = InBetweenDaysInclusive(new DateTime(2010,22),DayOfWeek.Sunday);// result == true; Oct 24,2010 is a Sundayresult = InBetweenDaysInclusive(new DateTime(2010,24),DayOfWeek.Sunday);// result == false; Oct 25,2010 is a Mondayresult = InBetweenDaysInclusive(new DateTime(2010,25),DayOfWeek.Sunday);
谢谢!
解决方法 根据开始和结束日期之间的差异是负数还是正/零,此函数将需要两个独立的分支.我可以完全脱离基地,但我认为这适用于所有情况:
// No state in method,so made it staticpublic static bool InBetweenDaysInclusive(DateTime date,DayOfWeek start,DayOfWeek end){ DayOfWeek curDay = date.DayOfWeek; if (start <= end) { // Test one range: start to end return (start <= curDay && curDay <= end); } else { // Test two ranges: start to 6,0 to end return (start <= curDay || curDay <= end); }}
作为参考,当我运行它时,您的测试数据返回以下内容,并为每个结果添加了Console.Writeline:
TrueTrueTrueFalse
编辑:我最后的解释太模糊了.这是一个固定的.
诀窍在于,开始,那么你有两个有效的范围:开始上限和下限结束.这将导致(开始< = curDay&& curDay< = upperBound)|| curDay< = end&&& lowerBound< = curDay) 然而,由于它们是边界,所以curDay总是< = upperBound和> = lowerBound,因此我们省略了该代码.
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